Question 1086213
<pre>
Here is a different approach that uses calculus, 
in case that's what you're studying. 

Let the equation of the circle be:

{{{(x-h)^2+(y-k)^2=r^2}}}

Substitute the two given points for (x,y)
 
{{{(6-h)^2+(2-k)^2=r^2}}}
{{{(3-h)^2+(-3-k)^2=r^2}}}

Subtracting the two equations:

 {{{(6-h)^2-(3-h)^2+(2-k)^2-(-3-k)^2=0}}}
{{{(6-h)^2-(3-h)^2=(-3-k)^2-(2-k)^2}}}

Factoring both sides as the difference of squares:

{{{((6-h)^""-(3-h)^"")((6-h)^""+(3-h)^"")=((-3-k)^""-(2-k)^"")((-3-k)^""+(2-k)^"")}}}
{{{(6-h^""-3+h^"")(6-h^""+3-h^"")=(-3-k^""-2+k^"")(-3-k^""+2-k^"")}}}
{{{(3^"")(9-2h^"")=(-5^"")(-1-2k^"")}}}
{{{27-6h=5+10k}}}
{{{22=6h+10k}}}
{{{11=3h+5k}}}

Here come da calculus:
 
The derivative (slope) of the line must be
the same as the derivative of the circle at 
the point of tangency
 
 {{{x-4y-15=0}}} is
 {{{1-4expr(dy/dx)=0}}}
 {{{-4expr(dy/dx)=-1}}}
 {{{dy/dx=1/4}}}

Let the circle have the equation

 {{{(x-h)^2+(y-k)^2=r^2}}}

Its derivative (by implicit differentiation) is

{{{2(x-h)+2(y-k)expr(dy/dx)=0}}}

Divide through by 2

{{{(x-h)+(y-k)expr(dy/dx)=0}}}

We substitute the point of tangency (3,-3) and the
derivative = 1/4: 

{{{(3-h)+(-3-k)expr(1/4)=0}}}
{{{4(3-h)+(-3-k)=0}}}
{{{12-4h-3-k=0}}}
{{{9=4h+k}}}

So we have the system of equations:

{{{system(11=3h+5k,9=4h+k)}}}

Solve the 2nd for k: 9-4h=k.  Substitute in 1st:

 11=3h+5(9-4h)
 11=3h+45-20h
-34=-17h
  2=h
  9=4h+k
  9=4(2)+k
  9=8+k
  1=k

So the center is (h,k) = (2,1)

To find the radius r (all you need for 
the equation is rē), substitute in

{{{(3-h)^2+(-3-k)^2=r^2}}}
{{{(3-2)^2+(-3-1)^2=r^2}}}
{{{(1)^2+(-4)^2=r^2}}}
{{{1+16=r^2}}}
{{{17=r^2}}} 

So the equation 
 
{{{(x-h)^2+(y-k)^2=r^2}}}

becomes:
 
{{{(x-2)^2+(y-1)^2=17}}}

{{{drawing(400,400,-8,12,-10,10,
circle(3,-3,.1),locate(3,-3,"(3,-3)"),
circle(6,2,.1),locate(6,2,"(6,2)"),
circle(2,1,.1),locate(2,1,"(2,1)"),
graph(400,400,-8,12,-10,10), circle(2,1,sqrt(17)),
line(-13,-7,15,0) )}}}

Edwin</pre>